Bicycle Rider’s Air Resistance In Example 3.9, we made the order-of-magnitude approximation that a person can produce 100–200 W of power while exercising. Based on the upper value of 200 W, estimate the speed at which a person can ride a bicycle at that level of exertion and still overcome air

Question

Bicycle Rider’s Air Resistance

In Example 3.9, we made the order-of-magnitude approximation that a person can produce 100–200 W of power while exercising. Based on the upper value of 200 W, estimate the speed at which a person can ride a bicycle at that level of exertion and still overcome air resistance (Figure 6.23). Express your answer in the dimensions of mph. In the calculation, neglect rolling resistance between the bicycle’s tires and the road, as well as the friction in the bearings, chain, and sprockets. A mathematical expression for power is P = Fv, where F is a force’s magnitude and v is the speed of the object to which the force is applied.

Approach
To find the speed, we assume that the only resistance that the rider encounters is air drag. The drag force is given by Equation (6.14),

[latex]F_{D}=\frac{1}{2} \rho Av^2C_{D} [/latex] (6.14)

and Table 6.4 lists

Table 6.4
Numerical Values of the Drag Coefficient and Frontal Area for Different Systems

	Frontal Area, A
System	[latex]ft^2[/latex]	[latex]m^2[/latex]	Drag Coefficient, [latex]C_{D}[/latex]
Economy sedan (60 mph)	20.8	1.9	0.34
Sports car (60 mph)	22.4	2.1	0.29
Sport-utility vehicle (60 mph)	29.1	2.7	0.45
Bicycle and rider (racing)	4.0	0.37	0.9
Bicycle and rider (upright)	5.7	0.53	1.1
Person (standing)	6.7	0.62	1.2

[latex]C_{D}= 0.9[/latex] with a frontal area of [latex]A = 4.0 ft^2[/latex] for a cyclist in racing position. To calculate the drag force, we will need the numerical value for the density of air, which is given as [latex]2.33 × 10^{-3} slugs/ft^3[/latex] in

Table 6.1.

Table 6.1 Density and Viscosity Values for Several Gases and Liquids at Room Temperature and Pressure

	Density,ρ		Viscosity, μ
Fluid	[latex]kg/m^3[/latex]	[latex]slug/ft^3[/latex]	kg/(m . s)	slug/(ft . s)
Air	1.20	[latex]2.33×10^{-3}[/latex]	[latex]1.8×10^{-5}[/latex]	[latex]3.8×10^{-7}[/latex]
Helium	0.182	[latex]3.53×10^{-4}[/latex]	[latex]1.9×10^{-5}[/latex]	[latex]4.1×10^{-7}[/latex]
Freshwater	1000	1.94	[latex]1.0×10^{-3}[/latex]	[latex]2.1×10^{-5}[/latex]
Seawater	1026	1.99	[latex]1.2×10^{-3}[/latex]	[latex]2.5×10^{-5}[/latex]
Gasoline	680	1.32	[latex]2.9×10^{-4}[/latex]	[latex]6.1×10^{-6}[/latex]
SAE 30 oil	917	1.78	0.26	[latex]5.4×10^{-3}[/latex]

Accepted Answer

We first will obtain a general symbolic equation for the cyclist’s speed, and then we will substitute numerical values into it. The power produced by the rider balances the loss from air resistance

[latex]P=\left(\frac{1}{2} ho Av ^2C_D ight) v \longleftarrow \left[P=Fv ight] [/latex]

[latex]=\frac{1}{2} ho Av ^3C_D[/latex]

The rider’s velocity becomes

[latex]v=\sqrt[3]{\frac{2P}{ ho AC_D} } [/latex]

Next, we will substitute numerical values into this equation. Converting the cyclist’s power to dimensionally consistent units with the factor listed in Table 3.6,

Table 3.6 Conversion Factors between Certain Quantities in the USCS and SI

Quantity	Conversion
Length	[latex]1 in. = 25.4 mm[/latex]
	[latex]1 in. = 0.0254 m[/latex]
	[latex]1 ft = 0.3048 m[/latex]
	[latex]1 mi = 1.609 km[/latex]
	[latex]1 mm = 3.9370 ×10^{-2} in.[/latex]
	[latex]1 m = 39.37 in.[/latex]
	[latex]1 m = 3.2808 ft[/latex]
	[latex]1 km = 0.6214 mi[/latex]
Area	[latex]1 in^2 = 645.16 mm^2[/latex]
	[latex]1 ft^2 = 9.2903 × 10^{-2} m^2[/latex]
	[latex]1 mm^2 = 1.5500 × 10^{-3 }in^2[/latex]
	[latex]1 m^2 = 10.7639 ft^2[/latex]
Volume	[latex]1 ft^3 = 2.832 × 10^{-2} m^3[/latex]
	[latex]1 ft^3 = 28.32 L[/latex]
	[latex]1 gal = 3.7854 × 10^{-3} m^3[/latex]
	[latex]1 gal = 3.7854 L[/latex]
	[latex]1 m^3 = 35.32 ft^3[/latex]
	[latex]1 L = 3.532 × 10^{-2} ft^3[/latex]
	[latex]1 m^3 = 264.2 gal[/latex]
	[latex]1 L = 0.2642 gal[/latex]
Mass	[latex]1 slug = 14.5939 kg[/latex]
	[latex]1 lbm = 0.45359 kg[/latex]
	[latex]1 kg = 6.8522 × 10^{-2} slugs[/latex]
	[latex]1 kg = 2.2046 lbm[/latex]
Force	1 lb = 4.4482 N
Force	1 N = 0.22481 lb
Pressure or stress	[latex]1 psi = 6895 Pa[/latex]
	[latex]1 psi = 6.895 kPa[/latex]
	[latex]1 Pa = 1.450 × 10^{-4 }psi[/latex]
	[latex]1 kPa = 0.1450 psi[/latex]
Work, energy, or heat	1 ft · lb = 1.356 J
	1 Btu = 1055 J
	1 J = 0.7376 ft · lb
	[latex]1 J = 9.478 × 10^{-4} Btu[/latex]
Power	1 (ft · lb)/s = 1.356 W
	1 hp = 0.7457 kW
	1 W = 0.7376 (ft · lb)/s
	1 kW = 1.341 hp

[latex]P=(200W) \left(0.7376\frac{(ft.lb)/s}{W} ight) [/latex]

[latex]=147.5(\cancel{W})\left(\frac{(ft.lb)/s}{\cancel{W}} ight)[/latex]

[latex] =147.5 \frac{ft.lb}{s}[/latex]

the cyclist’s speed becomes

[latex]v =\sqrt[3]{\frac{2(147.5(ft.lb)/S)}{(2.33 imes 10^{-3}slugs/ft^3)(4.0ft^2)(0.9)} } \longleftarrow \left[v =\sqrt[3]{\frac{2P}{ ho A C_{D}} } ight] [/latex]

[latex]=32.77\sqrt[3]{\frac{lb.ft^2.\cancel{ft^2}}{slug.\cancel{ft^2}.S} } [/latex]

[latex]=32.77\sqrt[3]{\frac{((\cancel{slug}.ft)/S^2).ft^2}{\cancel{slug}.S} } [/latex]

[latex]=32.77\sqrt[3]{\frac{ft^3}{S^3} } [/latex]

[latex]=32.77\frac{ft}{S} [/latex]

Finally, we convert this value to the conventional units of mph:

[latex]v =\left(32.77\frac{ft}{S} ight)\left(\frac{1}{5280}\frac{mi}{ft} ight)\left(3600\frac{S}{h} ight) [/latex]

[latex]=22.34\left(\frac{\cancel{ft}}{\cancel{S}} ight)\left(\frac{mi}{\cancel{ft}} ight)\left(\frac{\cancel{S}}{h} ight) [/latex]

[latex]= 22.34 mph[/latex]

Discussion
We recognize that this calculation overstates the rider’s speed because other forms of friction have been neglected in our assumptions. Nevertheless, the estimate is quite reasonable, and, interestingly, the resistance that air drag provides is significant. The power required to overcome air drag increases with the cube of the cyclist’s speed. If the rider exercises twice as hard, the speed will increase only by a factor of [latex]\sqrt[3]{2} ≈[/latex] 1.26 — 26 % faster.

[latex]v= 22.34 mph[/latex]

Question 6.8: Bicycle Rider’s Air Resistance In Example 3.9, we made the o...

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